The natural map $R\mapsto R[t]/(ft-1)$ is injective Let $R$ be a commutative ring and $f\in R$ such that $f$ is not a zero divisor (thanks Darij for correction). How can I rigorously show that 

The canonical map $R\mapsto R[t]/(ft-1)$ is injective.

Context: This step shows up in Rabinowitch trick (when proving Nullstellensatz) where $R=k[x_1, x_2, …, x_n]$. 
 A: If $\,r\in R\,$ lies in the kernel, then $\,r = (1-ft)\,g\,$ for some $\,g\in R[t],\, \deg g =: n.\,$ Then
$$\begin{eqnarray}    r &=&  (1-ft)\,g &\Rightarrow&\   g(0) = r\qquad\ \, {\rm via\ \  coef}\ t^0 \\[.2em]
 \Rightarrow\  (1\!+\!ft\!+\dots+\!(ft)^n)\, r &=&  (\color{#c00}{1\!-\!(ft)^{n+1}})\, g &\Rightarrow&\   g(0)\,f^{n+1}\! = 0\ \ {\rm via\ \  coef}\ t^{n+1} \\[.2em]
& & &\Rightarrow&\ \ \ \ {\color{#0a0}{r\ f^{n+1} = 0}}\\[-1em]
\!\!\text{by scaling by}\,\  1\!+\!ft\!+\dots+\!(ft)^n \,\  &\!\!=&  \dfrac{\color{#c00}{1\!-\!(ft)^{n+1}}}{\!\!\!\!\!\!\!\!\!1-ft}
\end{eqnarray}$$
$\color{#0a0}{\rm Therefore}\,$ if $\,\color{#0a0}f\,$ is not a zero-divisor then $\,\color{#0a0}{r = 0}\,$ and the map is injective. $\ $ QED
Remark $\ $  This is used in the computation of the kernel of localization maps. As I mentioned here, this simple proof seems to have been missed by most authors (e.g. Rotmans' Advanced Modern Algebra and all others I've seen give a proof like that in egreg's answer).
Edit $ $ Since the point of this approach seems to not be clear to some readers (cf. comments and downvotes), I will emphasize it explicitly. We do not use any recursion or telescopy above. Rather, we simply invoke the well-known identity $\,(1−x)(1+\cdots +x^n)=1−x^{n+1}\,$ for $\,x=ft.\,$ A common proof in many textbooks (and egreg's answer) amounts to repeating a telescopic proof of this identity in the proof at hand, vs. invoking it as a lemma as we do above (i.e. calling it by value vs. name in CS language). Inlining the proof (vs abstracting it out) only serves to complicate matters. Of course instances of the Factor Theorem (like the identity above) are ubiquitous in ring theory. We should strive to exploit this by appropriately abstracting out their application, esp. in instances like this where this structure is slightly hidden so it may not be immediately clear.
See also this answer for motivational conceptual background.
A: An element $a$ belongs to the kernel of the canonical morphism if and only if $a=(ft-1)P(t)$, where $P(t)\in R[t]$. Suppose $a\ne0$ and $P(t)=r_0+r_1t+\dots+r_nt^n$ exists, with $r_n\ne0$. Then
$$
(ft-1)P(t)=
-r_0+(fr_0-r_1)t+(fr_1-r_2)t^2+\dots+(fr_{n-1}-r_n)t^n+fr_nt^{n+1}
$$
so
\begin{align}
r_0&=-a\\
r_1&=-fa\\
r_2&=-f^2a\\
&\;\;\vdots\\
r_n&=-f^na\\
0&=f^{n+1}a
\end{align}
If $f$ is not a zero divisor, we have a contradiction.
Note that if $f$ is a zero divisor, the statement is false: if $fb=0$, then taking $P(t)=-b$, we have
$$
(ft-1)(-b)=fbt+b=b
$$
so $b$ belongs to the ideal generated by $ft-1$.
